We follow the proof of Proposition 3.2 in [1]}. In our particular case,
\(D\psi (x)={\left\lbrace \begin{array}{ll}2x \qquad \qquad &\text{if $|x|\le 1$}\\2\frac{x}{|x|} \qquad \qquad &\text{if $|x|> 1$,}\end{array}\right.}\qquad \text{and}\qquad D^2\psi (x)={\left\lbrace \begin{array}{ll}2 \qquad \qquad &\text{if $|x|\le 1$}\\0 \qquad \qquad &\text{if $|x|> 1$.}\end{array}\right.}\) 
We get
\(\begin{split}\mathcal {L}^{\nu _h}[\psi ](0)&=\int _{|z|\le 1}\int _0^1(1-t)z^TD^2\psi (tz)z\,\mathrm {d}t\,\mathrm {d}\nu _h(z)+\int _{|z|>1}\int _0^1D\psi (tz)\cdot tz\,\mathrm {d}t\,\mathrm {d}\nu _h(z)\\&=\int _{|z|\le 1}\int _0^1(1-t)z^TD^2\psi (tz)z\,\mathrm {d}t\,\mathrm {d}\nu _h(z)+\int _{1<|z|\le 1/t}\int _0^1D\psi (tz)\cdot tz\,\mathrm {d}t\,\mathrm {d}\nu _h(z)\\&\quad +\int _{|z|>1/t}\int _0^1D\psi (tz)\cdot tz\,\mathrm {d}t\,\mathrm {d}\nu _h(z)\\&=\int _{|z|\le 1}|z|^2\,\mathrm {d}\nu _h(z)+\int _{1<|z|\le 1/t}\int _0^12t^2|z|^2\,\mathrm {d}t\,\mathrm {d}\nu _h(z)+\int _{|z|>1/t}\int _0^12t|z|\,\mathrm {d}t\,\mathrm {d}\nu _h(z)\\&\ge \int _{|z|\le 1}|z|^2\,\mathrm {d}\nu _h(z)+\int _{1<|z|\le 1/t}\int _0^12t^2\,\mathrm {d}t|z|\,\mathrm {d}\nu _h(z)+\int _{|z|>1/t}\int _0^12t\,\mathrm {d}t|z|\,\mathrm {d}\nu _h(z)\\&\ge \int _{|z|>0}\min \lbrace |z|^2,|z|\rbrace \,\mathrm {d}\nu _h(z).\end{split}\) 
Hence,
\(\begin{split}\sup _{0<h<1}\int _{|z|>0}\min \lbrace |z|^2,|z|\rbrace \,\mathrm {d}\nu _h(z)&\le \sup _{0<h<1}\mathcal {L}^{\nu _h}[\psi ](0)\\&\le \Vert \mathcal {L}^\mu [\psi ]\Vert _{L^\infty (\mathbb {R}^N)}+\sup _{0<h<1}|(\mathcal {L}^{\nu _h}-\mathcal {L}^\mu )[\psi ](0)|,\end{split}\) 
which is finite due to the pointwise convergence and the assumption (REF ).
